Answer:
A. No effect
B. Results in the formation of additional hydrogen gas
C. Results in the formation of additional hydrogen gas
D. Results in the formation of additional hydrogen gas
E. No effect
F. No effect
Explanation:
The equilibrium in this question is
C(s) + H₂O (g) ⇄ CO(g) + H₂ (g)
and
Kp = pCO x pH₂/ pH₂O
where pCO, pH₂O and pH₂O are the partial pressures of CO, H₂ and H₂O.
We call the equilibrium constant Kp since only gases intervene in the expression for the constant.
A. adding more C to the reaction mixure
Adding more carbon which is a solid does not alter the pressure equilibrium constant, therefore, it has no effect on the equilibrium and consequently no effect on the quantity of hydrogen gas.
B. adding more H₂O to the reaction mixture
We can answer this part by using Le Chatelier's principle which states that a system at equilibrium will respond to a stress in such a way as to minimize the stress, hence restoring equilbrium.
One of the three possible stresses is an increase of reactant as in this case. The system will react by decreasing some of the added water. Thus the equilbrium shifts to the product side which will result in the formation of more hydrogen gas.
The difference of this part with respect to part A is that indeed the water gas is included in the equilibrium constant expression.
C. raising the temperature
This is another stress we can subject an equilibrium.
We are told the reaction is endothermic which means in going from left to right it consumes heat. Thus the equilibrium will shift to the product side by consuming some of the added heat favoring the production of more hydrogen gas.
D. increasing the volume of the reaction mixture
This the last of the stresses .
Increasing the volume of the reaction effectively decreases the pressure ( volume is inversely proportional to pressure ) so the equilibrium will shift to the side that has more pressure which is the product side: we have two moles of gases products vs. 1 mol gas in the reactant side.
Therefore, the equilibrium will shift to the right increasing the quantity of H₂.
E. adding a catalyst to the reaction mixture
The addition of a catalyst does not have an effect on the equilibrium constant. The catalyst will speed both the forward and reverse reaction decreasin the time to attain equilibrium.
So there is no effect on the quantity of H₂.
F. Adding an inert gas to reaction mixture
Assuming the volume of the reaction mixture remains constant, and we are not told such change in volume occurred, the addition of an inert gas does not have an effect in our equilibrium. The inert gasdoes not participate in the calculation for Kp.
The situation will be different if the volume of the reaction is allowed to increase, but again this is not stated in the question.
